Case Study Practice in Year 7 Edexcel Maths | Year 7 Edexcel 数学：案例分析实战演练

📚 Case Study Practice in Year 7 Edexcel Maths | Year 7 Edexcel 数学：案例分析实战演练

Real-life problems help you apply maths skills in practical situations. This article presents ten case studies that cover key topics in the Year 7 Edexcel curriculum, including numbers, fractions, percentages, geometry, and statistics. Each case study encourages you to think critically and use step-by-step reasoning.

现实生活中的问题有助于你将数学技能应用到实际情境中。本文提供了十个案例研究，涵盖 Year 7 Edexcel 课程的关键主题，包括数字、分数、百分比、几何和统计。每个案例鼓励你进行批判性思考，并使用逐步推理。

1. Shopping on a Budget | 预算购物

You have £50 to buy school supplies. A notebook costs £2.50, a pen costs £1.20, and a ruler costs £0.80. You need 5 notebooks, 10 pens, and 3 rulers. Do you have enough money, and how much will be left?

你有50英镑购买学习用品。一本笔记本2.50英镑，一支笔1.20英镑，一把尺子0.80英镑。你需要5本笔记本、10支笔和3把尺子。你有足够的钱吗？会剩下多少钱？

Calculate the total cost: 5 × £2.50 = £12.50, 10 × £1.20 = £12.00, 3 × £0.80 = £2.40. Sum these to get £12.50 + £12.00 + £2.40 = £26.90.

计算总费用：5 × 2.50英镑 = 12.50英镑，10 × 1.20英镑 = 12.00英镑，3 × 0.80英镑 = 2.40英镑。加起来得到 12.50英镑 + 12.00英镑 + 2.40英镑 = 26.90英镑。

Since £26.90 is less than £50, you have enough money. Subtract to find the change: £50.00 – £26.90 = £23.10 left over.

因为26.90英镑小于50英镑，所以钱足够了。减去后得到找零：50.00英镑 – 26.90英镑 = 剩下23.10英镑。

2. Planning a Day Out | 计划一日游

A family plans to visit a theme park. They leave home at 08:45 and the drive takes 1 hour 25 minutes. The park closes at 17:30. They want to stay for at least 6 hours. Is it possible, and what time should they leave the park to get home by 19:00 if the return journey takes the same time?

一个家庭计划去主题公园玩。他们8:45离开家，车程需要1小时25分钟。公园17:30关门。他们想至少玩6小时。这可能吗？如果回程时间相同，他们应该什么时间离开公园才能在19:00前到家？

Arrival time: 08:45 + 1 hour 25 min = 10:10. From 10:10 to 17:30 is 7 hours 20 minutes, which is longer than 6 hours, so they can enjoy the park.

到达时间：08:45 + 1小时25分钟 = 10:10。从10:10到17:30是7小时20分钟，比6小时长，所以他们可以在公园里尽兴。

To be home by 19:00, subtract the journey time: 19:00 – 1 hour 25 min = 17:35. They must leave the park by 17:35.

为了在19:00前到家，减去路程时间：19:00 – 1小时25分钟 = 17:35。他们必须在17:35前离开公园。

However, the park closes at 17:30, so they can leave just before closing and still arrive home around 18:55. It works perfectly!

然而，公园17:30关门，所以他们可以在关门前离开，大约18:55到家。完美！

3. Scaling a Recipe | 调整食谱分量

A recipe for 4 people requires 2½ cups of flour, ¾ cup of sugar, and 1&frac13; teaspoons of vanilla. You need to serve 6 people. How much of each ingredient do you need?

一份供4人食用的食谱需要 2½ 杯面粉、¾ 杯糖和 1&frac13; 茶匙香草精。你需要为6人备餐。每种原料需要多少？

Find the multiplier: 6 ÷ 4 = 1.5. Multiply each quantity by 1.5.

找出倍数：6 ÷ 4 = 1.5。将每种数量乘以1.5。

Flour: 2½ = 5/2, so 5/2 × 1.5 = 5/2 × 3/2 = 15/4 = 3¾ cups.

面粉：2½ = 5/2，所以 5/2 × 1.5 = 5/2 × 3/2 = 15/4 = 3¾ 杯。

Sugar: ¾ × 1.5 = ¾ × 3/2 = 9/8 = 1&frac18; cups.

糖：¾ × 1.5 = ¾ × 3/2 = 9/8 = 1&frac18; 杯。

Vanilla: 1&frac13; = 4/3, so 4/3 × 1.5 = 4/3 × 3/2 = 12/6 = 2 teaspoons.

香草精：1&frac13; = 4/3，所以 4/3 × 1.5 = 4/3 × 3/2 = 12/6 = 2 茶匙。

4. Garden Design | 花园设计

A rectangular lawn measures 8 m by 5 m. A 1 m wide flower bed is added along one long side and one short side, forming an L-shape. Find the area of the flower bed and the remaining lawn area.

一块长方形草坪尺寸是8米乘5米。沿着一条长边和一条短边修建一个1米宽的花坛，形成L形。求花坛的面积和剩余草坪的面积。

First, total garden area after adding beds: the overall dimensions become (8+1) m by (5+1) m = 9 m by 6 m. Total area = 9 × 6 = 54 m².

首先，添加花坛后的整体尺寸变为 (8+1) 米乘 (5+1) 米 = 9米乘6米。总面积 = 9 × 6 = 54 平方米。

Original lawn area = 8 × 5 = 40 m². So the flower bed area = 54 – 40 = 14 m².

原草坪面积 = 8 × 5 = 40 平方米。所以花坛面积 = 54 – 40 = 14 平方米。

Alternatively, calculate the L-shaped bed as two rectangles: (9 m × 1 m) + (5 m × 1 m) = 9 + 5 = 14 m² (be careful not to double count the corner).

或者，将L形花坛作为两个矩形计算：(9米 × 1米) + (5米 × 1米) = 9 + 5 = 14 平方米（注意不要重复计算转角）。

5. Classroom Survey | 课堂调查

Students recorded the number of books they read in a month: 3, 4, 2, 5, 3, 4, 6, 2, 3, 5. Calculate the mean, mode, and range. Also suggest a suitable chart to display the data.

学生们记录了一个月内阅读的书的数量：3、4、2、5、3、4、6、2、3、5。计算平均数、众数和极差。并建议一种合适的图表来展示数据。

Order the data: 2, 2, 3, 3, 3, 4, 4, 5, 5, 6. Mean = sum ÷ 10 = (2+2+3+3+3+4+4+5+5+6) ÷ 10 = 37 ÷ 10 = 3.7 books.

将数据排序：2、2、3、3、3、4、4、5、5、6。平均数 = 总和 ÷ 10 = (2+2+3+3+3+4+4+5+5+6) ÷ 10 = 37 ÷ 10 = 3.7 本书。

Mode: the most frequent value is 3 (appears three times). Range = maximum – minimum = 6 – 2 = 4.

众数：出现次数最多的值是3（出现三次）。极差 = 最大值 – 最小值 = 6 – 2 = 4。

A bar chart or a frequency table would be suitable to display how many students read each number of books.

柱状图或频率表适合展示有多少学生读了相应数量的书。

6. Discounts at the Store | 商店折扣

A jacket originally costs £60. During a sale, there is a 25% discount. Later, an extra 10% off is applied to the reduced price. What is the final price, and what is the overall percentage saving compared to the original?

一件夹克原价60英镑。在促销期间，打七五折（25%折扣）。之后，在减价的基础上再打九折（额外10% off）。最终价格是多少？与原价相比，总共节省了百分之几？

First discount: 25% off £60 = 0.25 × 60 = £15 reduction. Price after first discount = £60 – £15 = £45.

第一次折扣：60英镑的25% off = 0.25 × 60 = 减少15英镑。第一次折扣后价格 = 60 – 15 = 45英镑。

Second discount: 10% off £45 = 0.10 × 45 = £4.50 reduction. Final price = £45 – £4.50 = £40.50.

第二次折扣：45英镑的10% off = 0.10 × 45 = 减少4.50英镑。最终价格 = 45 – 4.50 = 40.50英镑。

Overall saving = £60 – £40.50 = £19.50. Overall percentage saving = (£19.50 ÷ £60) × 100% = 32.5%.

总共节省：60 – 40.50 = 19.50英镑。总节省百分比 = (19.50 ÷ 60) × 100% = 32.5%。

7. Angles in a Maze | 迷宫中的角度

A robot moves through a maze using the commands: turn 60° right, turn 120° left, turn 90° right. Calculate the total clockwise turn and the robot’s final facing direction if it started facing north.

一个机器人在迷宫中按照指令移动：右转60°，左转120°，右转90°。计算机器人总的顺时针转动角度，以及如果它开始时面向北方，最终面向的方向。

Treat right turns as positive clockwise, left turns as negative clockwise (or anti-clockwise). Total clockwise turn = 60° – 120° + 90° = 30° clockwise.

将右转视为顺时针正向，左转视为顺时针负向（或逆时针）。总顺时针转动 = 60° – 120° + 90° = 顺时针30°。

Starting north (0°), a 30° clockwise turn makes the robot face north-east (specifically 30° from north towards east).

从正北（0°）开始，顺时针转30°使机器人面向东北方向（具体为从北向东偏30°）。

8. Packing Boxes | 打包盒子

A rectangular box measures 30 cm long, 20 cm wide, and 15 cm tall. Small cubes of side length 5 cm are packed inside without gaps. How many cubes fit? If each cube holds 2 kg, what is the total weight supported?

一个长方体的盒子长30厘米、宽20厘米、高15厘米。边长为5厘米的小立方体无缝隙地装入其中。可以装多少个立方体？如果每个立方体承重2千克，总共支撑的重量是多少？

Volume of box = 30 × 20 × 15 = 9000 cm³. Volume of one cube = 5 × 5 × 5 = 125 cm³.

盒子体积 = 30 × 20 × 15 = 9000 立方厘米。一个立方体体积 = 5 × 5 × 5 = 125 立方厘米。

Number of cubes = 9000 ÷ 125 = 72 cubes. Alternatively, along length: 30÷5 = 6, width: 20÷5 = 4, height: 15÷5 = 3, so 6 × 4 × 3 = 72.

立方体数量 = 9000 ÷ 125 = 72 个。或者，沿长边：30÷5 = 6，宽边：20÷5 = 4，高：15÷5 = 3，所以 6 × 4 × 3 = 72。

Total weight = 72 × 2 kg = 144 kg.

总重量 = 72 × 2 千克 = 144 千克。

9. Fraction Pizza Party | 分数披萨派对

Three friends share a pizza cut into 8 equal slices. Alex eats 3/8, Bella eats 1/4, and Chloe eats the rest. What fraction does Chloe eat, and who eats the most?

三个朋友分享一个被切成8等份的披萨。Alex吃了3/8，Bella吃了1/4，Chloe吃了剩下的部分。Chloe吃了多少？谁吃得最多？

Convert fractions to eighths: 1/4 = 2/8. Total eaten by Alex and Bella = 3/8 + 2/8 = 5/8.

将分数转化为八分之几：1/4 = 2/8。Alex和Bella总共吃了 3/8 + 2/8 = 5/8。

Chloe eats 1 – 5/8 = 3/8 of the pizza. Alex and Chloe both eat 3/8, while Bella eats 2/8, so Alex and Chloe eat the most, tied.

Chloe吃了 1 – 5/8 = 3/8 的披萨。Alex和Chloe都吃了3/8，Bella吃了2/8，所以Alex和Chloe吃得最多，并列第一。

If the pizza had been cut into 6 slices, can they still get these fractions? Not exactly, because 3/8 of 6 is not a whole number of slices; the sharing works best with denominators that match the number of slices.

如果披萨被切成6块，他们还能得到这些分数吗？不能精确，因为6的3/8不是整数片；当分母与总片数匹配时，分享才最方便。

10. Tile Patterns | 瓷砖图案

A floor pattern uses rows of square tiles. Row 1 has 3 tiles, Row 2 has 6 tiles, Row 3 has 9 tiles, and so on. How many tiles are in Row 10? Write a rule (formula) for the number of tiles in Row n.

一种地板图案使用成排的方形瓷砖。第1行有3块瓷砖，第2行有6块瓷砖，第3行有9块瓷砖，以此类推。第10行有多少块瓷砖？写出第n行瓷砖数量的规则（公式）。

Recognise the pattern: each row increases by 3 tiles. This is a multiplication sequence: row 1 = 3×1, row 2 = 3×2, row 3 = 3×3. So row n has 3n tiles.

识别规律：每行增加3块瓷砖。这是一个乘法序列：第1行 = 3×1，第2行 = 3×2，第3行 = 3×3。所以第n行有 3n 块瓷砖。

For row 10, substitute n = 10: 3 × 10 = 30 tiles.

对于第10行，代入 n = 10：3 × 10 = 30 块瓷砖。

If the first row had 5 tiles and each subsequent row added 4, the formula would be 5 + 4(n-1) = 4n + 1. Sequences can help predict tiling costs.

如果第一行有5块瓷砖，之后每行增加4块，公式将是 5 + 4(n-1) = 4n + 1。数列有助于预测铺砖成本。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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Case Study Practice in Year 7 Edexcel Maths | Year 7 Edexcel 数学：案例分析实战演练